1
Problem 1.2
2
Problem 1.3
3
Problem 1.4
4
5
Problem 1.6
Make a guess at the order of magnitude of the mass (e.g., 0.01, 0.1, 1.0, 10, 100, or 1000 lbm or kg) of standard air that is in a room 10 ft by 10 ft by 8 ft, and then compute this ma
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General Engineering Data
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General Engineering Data
Resistance Coefficient k, Equivalent Length l/d,
and Flow Coefficient Cv
Pressure loss test data for a wide variety of valves and fittings are
available from the work
Problem 3.1
D = 0.75 m. The gas is at an absolute pressure of 25 MPa and a temperature of 25C. What is the mass in the tank? If the maximum allowable wall stress in the tank is 210 MPa, find the minimum theoretical wall thickness of the tank. Given:
Problem 4.2
A mass of 3 kg falls freely a distance of 5 m before contacting a spring attached to the ground. If the spring stiffness is 400 N/m, what is the maximum spring compression?
Given: Data on mass and spring Find: Maximum spring compression
Problem 5.3
Problem 5.4
Problem 5.6
Problem 5.9
The x component of velocity in a steady incompressible flow field in the xy plane is u = Ax/(x2 + y2), where A = 10 m2/s, and x and y are measured in meters. Find the simplest y component of velocity
Problem 7.4
Problem 7.5
Nondimensionalizing the velocity, pressure, spatial measures, and time:
u* =
u V
p* =
p p
x* =
x L
r* =
r L
t* = t
V L
Hence
L t* V
u =V u*
p = p p *
x = Lx*
r = Dr*
t=
Substituting into the governing equati
Problem 8.1
Problem 8.2
Given: Data on air flow in duct Find: Volume flow rate for turbulence; entrance length Solution The given data is D = 0.25 m
2 -5 m
From Fig. A.3
= 1.46 10
s
The governing equations are Re = V D Recrit = 2300 Q = 4
Problem 9.1 (In Excel)
Solution Governing equations: The critical Reynolds number for transition to turbulence is Re crit = VL crit/ = 500000
The critical length is then L crit = 500000/V Tabulated or graphical data: = = 3.79E-07 0.00234 (Table A.
Problem 11.2
Given: Data on an air compressor Find: Whether or not the vendor claim is feasible Solution The data provided, or available in the Appendices, is: p1 = 101 kPa p2 = ( 650 + 101) kPa J cp = 1004 kg K T1 = ( 20 + 273) K T2 = ( 285 + 273
Problem 12.1
Problem 12.2
Problem 12.4 (In Excel)
Given: Find:
Data on flow in a passage Pressure at downstream location
Solution The given or available data is: R = k = T1 = p1 = V1 = M2 = Equations and Computations: From T 1 and Eq. 11.17 c1 =
Problem 6.4
Problem 6.5
Problem 6.6
Given: Velocity field Find: Expressions for local, convective and total acceleration; evaluate at several points; evaluat pressure gradient Solution The given data is A = 2 1 s = 1 1 s = 2 kg m
3
u = A x sin